Given
2Al2O3 (s) --> 4Al(s) + 3O2 (g) (standard enthalpy change= 3351.4 kJ)
a) What is the heat of formation of aluminum oxide?
How do I find heat of formation from standard enthalpy change? I know how to do it when I'm given enthalpy, but they aren't the same thing- correct?
Also, if this question gave me delta-H, I know that the heat of formation for 4Al = 0 and 3O2 = 0 because they are elementary species. This would leave me with 2Al2O3 = delta-H = x (whatever delta-H is) However, since I have 2 moles of 2Al2O3, would I have to divide delta-H by 2? Sorry if that was confusing.
The delta H for the reaction listed is just the reverse of the heat of formation. So add a negative sign to make it -3351 kJ for the reaction, then divide by 2 to make it 1/2*-3351 = 1676 kJ/mol.
-1676
Why did the heat of formation cross the road? To get divided by 2 and turn into a negative! But seriously, you're on the right track. The delta H for the reaction is indeed the negative of the heat of formation. So, just remember to add a negative sign to the standard enthalpy change value given (-3351.4 kJ) and then divide it by 2 to get the heat of formation per mole (1676 kJ/mol). Keep up the good work, and you'll ace those thermochemistry calculations!
The heat of formation of a compound is defined as the enthalpy change when one mole of the compound is formed from its constituent elements in their standard states.
In this case, the standard enthalpy change for the given reaction is -3351.4 kJ. Since the equation represents the formation of 2 moles of Al2O3, you need to divide the standard enthalpy change by 2 to find the heat of formation per mole of Al2O3.
Therefore, the heat of formation of aluminum oxide (Al2O3) is -3351.4 kJ / 2 = -1675.7 kJ/mol.
Note that the heat of formation is always given per mole of the compound formed, so there is no need to further divide the value by 2.
To find the heat of formation of aluminum oxide (Al2O3) from the given standard enthalpy change (-3351.4 kJ), you need to understand the relationship between heat of formation (ΔHf) and standard enthalpy change (ΔH).
The heat of formation (ΔHf) is the enthalpy change when one mole of a compound is formed from its constituent elements, with all substances in their standard states at a specified temperature and pressure. The standard enthalpy change (ΔH) is the difference in enthalpy between the products and reactants of a chemical reaction when all substances are at their standard states.
In this case, the given standard enthalpy change (-3351.4 kJ) represents the enthalpy change that occurs when 2 moles of aluminum oxide (2Al2O3) decompose to form 4 moles of aluminum (4Al) and 3 moles of oxygen gas (3O2).
Since the heat of formation is the reverse reaction of the given reaction, you need to reverse the sign of the given standard enthalpy change. So, the heat of formation of aluminum oxide (ΔHf of Al2O3) would be -(3351.4 kJ).
However, since the heat of formation is typically expressed per mole, and the given enthalpy change is for 2 moles of aluminum oxide, you need to divide the value by 2 to get the heat of formation per mole. Therefore, the heat of formation of aluminum oxide (ΔHf of Al2O3) would be (-3351.4 kJ) / 2 = -1675.7 kJ/mol.
Therefore, the heat of formation of aluminum oxide is approximately -1675.7 kJ/mol.